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Results tagged with galois-theory
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user 298680
Galois theory allows one to reduce certain problems in field theory, especially those related to field extensions, to problems in group theory. For questions about field theory and not Galois theory, use the (field-theory) tag instead. For questions about abstractions of Galois theory, use (galois-connections).
1
vote
Accepted
The restriction $\sigma|K$ is in $Gal(K/k)$
If $\sigma \in \mathrm{Gal}(KF/F)$, then
$$\sigma(x)=x \quad\forall x \in F$$
Since $k$ is a subset of $F$, it follows that
$$\sigma(x)=x \quad\forall x \in k$$
In particular, if $K/k$ is a finite …
- 23.9k
4
votes
Proof of $\mathrm{Gal}(K_1K_2K_3/K)\cong \mathrm{Gal}(K_1/K)\times \mathrm{Gal}(K_2/K)\times...
In order to have the isomorphism$\newcommand{\Gal}{\text{Gal}}$
$$\Gal(K_1K_2K_3/K)\cong \Gal(K_1/K)\times \Gal(K_2/K)\times \Gal(K_3/K),$$
you have to assume that each $K_i$ is linearly independent n …
- 23.9k
7
votes
1
answer
542
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Abelian extension and Dirichlet characters
$\newcommand\Q {\Bbb Q}
\newcommand\Z[1]{\Bbb Z/#1 \Bbb Z}
\newcommand\Gal{\mathrm{Gal}}
\newcommand\K[1]{\Bbb Q(\zeta_{#1})}$
Let $K/\Q$ be an abelian extension with Galois group $G$ and let $\rho : …
- 23.9k
1
vote
Accepted
Union of infinite subfields of the complex plane
The problem here is that the fields $K_n = \Bbb Q(\sqrt n)$ are not totally ordered for the inclusion.
Let $A$ to be the union of the $K_n$, for $n≥1$. Consider $x=\sqrt 2 \in A$ and $y = \sqrt 3 \in …
- 23.9k
3
votes
Accepted
Hilbert's Theorem 90 for infinite extensions
Let $L/K$ be an infinite (algebraic) Galois extension.
You are completely right about the Galois group $\mathrm{Gal}(L/K)$ being a profinite group. More precisely, $\mathrm{Gal}(L/K)$ is the inverse l …
- 23.9k
4
votes
Accepted
Why is $[\mathbb{F}_{p^n}:\mathbb{F}_{p}] = n$?
Let $d$ be the dimension of $\mathbb{F}_{p^n}$ over $\mathbb{F}_{p}$. Then, you have an isomorphism as vector spaces:
$$\mathbb{F}_{p^n} \cong \mathbb{F}_{p}^d$$
The cardinality of the first one is $p …
- 23.9k
4
votes
Accepted
The ring of integral elements of $L$ with $L/K$ Galois.
Yes, here $\mathcal O = \mathcal O_L$ denotes the set of all elements of $L$ that are integral over the ring of integers $\mathcal O_K$ of $K$ (for instance if $L/K$ is an extension of number fields). …
- 23.9k
3
votes
Accepted
Basic question in Galois theory (on applying elements of the Galois group to a root of polyn...
Hint 1 :
For your first question, notice that an element in $K$ is of the form $P(\theta)$, with $P(X) \in \Bbb Q[X]$ and that for any $g \in G$, $g(P(\theta)) = P(g(\theta))$.
Answer 1 :
- 23.9k
1
vote
Accepted
Galois group $G= G_1 \times G_2$ - corresponding fixed fields
As for the first equality, you can notice that
$$\mathrm{Gal}(K / K_1K_2) \subset G_1 \cap G_2.$$
For instance, if $\sigma \in \mathrm{Gal}(K/F)$ fixes $K_1 K_2$ (i.e. $\sigma \in \mathrm{Gal}(K / K_ …
- 23.9k
2
votes
Accepted
Is $K/F$ also Galois when $K/k$ is Galois and $k \subset F \subset K$?
Yes, this is true. This is proved in Lang, Algebra, combining theorems 3.4 and 4.5.
— On the one hand, $K/F$ is separable. Let $x \in K$ and let $f$ be the minimal polynomial of $x$ over $F$ (we only …
- 23.9k
1
vote
Accepted
Abelian extension and Dirichlet characters
$\newcommand\Q {\Bbb Q}\newcommand\Gal{\operatorname{Gal}}$Thanks to Mathmo123's comments, I can provide an answer.
I will show that $L(\rho,s)=L(\chi,s)$, by showing that
$$\det(1-\rho(\sigma_P)p^{- …
- 23.9k
11
votes
1
answer
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Totally real Galois extension of given degree
Let $n≥1$ be an integer. I would like to prove (or disprove) the existence of a subfield $K \subset \Bbb R$ such that $K/\Bbb Q$ is Galois and has degree $n$.
It is easy to construct such a subfi …
- 23.9k
3
votes
Accepted
Can I find a Galois extension which contains a finite set of algebraic elements?
Yes, this is possible if $K/F$ is a separable extension. Let $P_i \in K[X]$ the minimal polynomial of $a_i$ and define $P$ to be the product of all the $P_i$.
Moreover, let $L$ the splitting field of …
- 23.9k
8
votes
Accepted
Abelian Galois group of $f$ implies splitting is simple extensions by a root of $f$.
As John Martin pointed out, you can take any root $\alpha \in K$ of $f$ and then you show that $\Bbb Q(\alpha)$ is Galois over $\Bbb Q$.
In particular, this means that every root of the minimal polyn …
- 23.9k
2
votes
1
answer
402
views
"Frobenius element" in cyclotomic extension for ramified primes
Let $m = \prod\limits_{q \text{ prime}} q^{n(q)}\;$ be a positive integer, let $p$ be a prime factor of $m$.
$\def\Q {\Bbb Q}
\def\Gal{\mathrm{Gal}}
\newcommand\K[1]{\Bbb Q(\zeta_{#1})}$
Question:
…
- 23.9k